Question

A function f has derivatives of all orders for −1(c) Write the first four nonzero terms and the general term of the Maclaurin series for g(x) =∫{0,x} f(t) dt.

a) x - x²/2 + x³/3 - x⁴/4
b) x + x²/2 + x³/3 + x⁴/4
c) x - x²/4 + x³/9 - x⁴/16
d) x + x²/4 + x³/9 + x⁴/16

Answer 1

The correct Maclaurin series for g(x) = ∫{0,x} f(t) dt is found by integrating each term of f(x)'s Maclaurin series. The signs alternate, and the denominators increase by factorial terms, resulting in option (c): x - x²/4 + x³/9 - x´/16.

Step-by-step explanation:

To find the correct option for the Maclaurin series for the integral g(x) = ∫{0,x} f(t) dt, we first acknowledge that f(x) has derivatives of all orders at x = 0. The Maclaurin series for f(x) can be written as f(x) = f(0) + f'(0)x + f''(0)x²/2! + f'''(0)x³/3! + …. The integral of each term of f(x) with respect to x from 0 to x will give us the series for g(x).

To find the Maclaurin series for g(x), we integrate each term of the Maclaurin series of f(x):

• ∫ f(0)dx = f(0)x
• ∫ f'(0)xdx = f'(0)x²/2
• ∫ f''(0)x²/2! dx = f''(0)x³/3! x 3/2 = f''(0)x³/6
• ∫ f'''(0)x³/3! dx = f'''(0)x´/4! x 4/3 = f'''(0)x´/24

Each resulting term corresponds to the integral of the respective Maclaurin term. Considering the function f(x) is non-zero at x = 0, we would have a series that starts with the linear term x. Hence, the correct format of the terms would alternate signs due to the integration of even and odd powers of x. Therefore, the possible correct answer should have alternating signs and increasing factorial denominators, corresponding to option (c) x - x²/4 + x³/9 - x´/16.